A projectile is shot from the ground at an angle of #pi/12 # and a speed of #7 /18 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

3 Answers
Jul 19, 2017

distance from starting point is #5/1296m~~3.858"x"10^-3m#

Explanation:

Range, #R=V^2/gsin(2theta)#

When the particle reaches its maximum height, the distance from starting point will be #R/2#

#R/2=V^2/(2g)sin(2theta)# where V is the initial launching speed

#R/2=(7/18)^2/(2g)sin(2*pi/12)#

#R/2=(49/324)/(2g)sin(pi/6)#

[Take #g=9.8ms^-2#]

#R/2=(5/648)sin(pi/6)#

#R/2=5/1296m~~3.858"x"10^-3m#

Jul 19, 2017

#"distance" = 0.00389# #"m"#

Explanation:

We're asked to find the distance from the launch point a projectile is when it reaches its maximum height.

We can use the equation

#(v_y)^2 = (v_(0y))^2 - 2gh#

to find the maximum height #h#.

When the particle is at its maximum height, the instantaneous velocity #v_y# is #0#.

The initial #y#-velocity #v_(0y)# is

#v_(0y) = v_0sinalpha_0 = (7/18color(white)(l)"m/s")sin(pi/12) = 0.101# #"m/s"#

Plugging in known values, we have

#0 = (0.101color(white)(l)"m/s")^2 - 2(9.81color(white)(l)"m/s"^2)h#

#h = ((0.101color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(5.16xx10^-4# #color(red)("m"#

To find the horizontal distance covered #Deltax#, we use the equation

#Deltax = v_(0x)t#

#v_(0x) = v_0cosalpha_0 = 0.376# #"m/s"#

The time #t# when it reaches its maximum height is given by

#v_y = v_(0y) -g t#

Rearranging this equation and plugging in known values:

#t = (v_(0y)-v_y)/g = (0.101color(white)(l)"m/s" - 0)/(9.81color(white)(l)"m/s"^2) = 0.0103# #"s"#

Thus, we have

#Deltax = (0.376color(white)(l)"m/s")(0.0103color(white)(l)"s") = color(green)(0.00385# #color(green)("m"#

The distance from the starting point is thus

#r = sqrt((Deltax)^2 + h^2) = sqrt((color(green)(0.00385color(white)(l)"m"))^2 + (color(red)(5.16xx10^-4color(white)(l)"m"))^2)#

#= color(blue)(0.00389# #color(blue)("m"#

Jul 19, 2017

THe distance is #=0.0039m#

Explanation:

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=7/18*sin(1/12pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=7/18sin(1/12pi)-g*t#

#t=7/(18g)*sin(1/12pi)#

#=0.0103s#

The greatest height is

#h=(7/18sin(1/12pi))^2/(2g)=0.00052m#

Resolving in the horizontal direction #rarr^+#

To find the horizontal distance, we apply the equation of motion

#s=u_x*t#

#=7/18cos(1/12pi)*0.0103#

#=0.00387m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(0.00052^2+0.00387^2)#

#=0.0039m#